2.5 The function f(z) = 1z

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The real function f(x) = 1=x is a perfectly straightforward function which

is de_ned everywhere except, of course, at x = 0. Since you can do in C

2.5. THE FUNCTION F(Z) = 1

Z 59

everything that you can do in R, the function f(z) = 1=z must also make

sense except at z = 0.

We can say immediately that f(z) = 1=z = _z=z_z, so 1=z does two things:

_rst it takes the conjugate of z, and then it scales by dividing by the square

of the modulus. If this is 1, then the only e_ect is to reect z in the X-axis.

In order to make our lives easier, we decompose f into these two parts, the

inversion map inv(z) = z=z_z and the conjugation map_(z) = _z. and look at

these separately.

To start to get a grip on the inv function, notice that in polar form, (r; _)

gets sent to (1=r; _). A point on the unit circle will stay _xed, points on the

axes stay on the axes. The origin gets sent o_ to in_nity, points close to the

origin get sent far away but preserve the angle. If we take the unit square in

the plane, the point _ 1

1 _ gets sent to _ 0:5

0:5 _.

The top edge of the unit square, y = 1;0 _ x _ 1, gets sent to a curve joining

_ 0:5

0:5 _ _ 0

1 _ which is left _xed by the map as it lies on the unit circle. The

equation of the curve is given by

u = x=(x2 + 1); v = 1=(x2 + 1)

which can be written as

u = vp1=v 􀀀 1

The right edge behaves similarly.

The left edge is sent to the Y-axis for values greater than 1, the bottom edge

to the X-axis with values from 1 to in_nity.

The before and after pictures are _gure 2.17 and _gure 2.18 respectively.

Note that the point density is greatest closest to the origin. You should be

able to see why this is so. (Hint: think of the derivative of 1=r.)

If we take a disk, it can be discovered experimentally that the image is also

a disk in most cases. Some before and after pictures are _gure 2.19 and

_gure 2.20 respectively.

This is a harder one to calculate:

60 CHAPTER 2. EXAMPLES OF COMPLEX FUNCTIONS

Figure 2.17: The Unit Square

Figure 2.18: The Inversion of the unit square

2.5. THE FUNCTION F(Z) = 1

Z 61

Figure 2.19: A disk

Figure 2.20: The Inversion of the disk

62 CHAPTER 2. EXAMPLES OF COMPLEX FUNCTIONS

Exercise 2.5.1 Can you _nd an expression for the inversion of the boundary

of the disk?

If you do some experimenting with a program that does inversions, you will

discover that it looks very much as if the inversion of a circle is a circle

except in the degenerate case where the circle passes through the origin.

This is indeed the case.

In order to see this, write the circle with centre _ a

b _ and radius R in polar

coordinates to get the equation

r2

􀀀 2ar cos _ 􀀀 2br sin _ = R2

􀀀 a2

􀀀 b2

Now the angle is unchanged, so the inversion is the set of (s; _) satisfying

1=s2

􀀀 2a cos _=s 􀀀 2b sin _=s = R2

􀀀 a2

􀀀 b2

which after some rearrangement gives

s2

􀀀 2a cos _=(a2 + b2

􀀀 R2) 􀀀 2b sin _=(a2 + b2

􀀀 R2) = 􀀀1=((a2 + b2

􀀀 R2)

This is a circle with centre at _ a=(a2 + b2 􀀀 R2)

b=(a2 + b2 􀀀 R2) _ and radius a rather horrible

value which can be written down with some patience. If the original

circle goes through the origin, the radius of the inverted circle is in_nite, and

its centre is also shifted o_ to in_nity. This actually gives a straight line.

Exercise 2.5.2 Verify the claim that the equation degenerates to a straight

line when R2 = a2 + b2.

Sneaky Alternative Methods

There is a somewhat neater way of proving that inversions take circles to

circles; it requires that we _nd a way of describing circles which is di_erent

from the usual one.

Suppose we write

jz 􀀀 aj

jz 􀀀 bj

= r

for some positive real r, and complex a; b. If r = 1 this just gives the straight

line bisecting the line segment from a to b. If r 6= 1, it gives a circle cutting

2.5. THE FUNCTION F(Z) = 1

Z 63

the line segment between a and b and it is easy to write down its equation

in more standard forms. Also, any circle can be written in this form.

Now putting w = 1=z in this equation and doing a bit of messing around

with algebra gives a new equation in the same form. Which is also a circle,

or maybe a straight line.

Exercise 2.5.3 Do the algebra to show that the representation is really that

of a circle (or straight line if r = 1).

Do the algebra to show that w = 1=z in this equation gives a new circle (or

possibly a straight line).

Exercise 2.5.4 Show that the unit circle can be represented in the sneaky

form with r = 2. Show that any circle can be written in this form with r = 2.

Another way of representing any circle is in the form

A _ Az_z + Bz + _B

_z + C _ C = 0

for complex numbers A;B;C.

If A = 0 this is a straight line, if C = 0 it passes through the origin.

For this form also, it is easy to con_rm that inversion takes circles to circles,

where a straight line is just a rather extremal case of a circle. Malcolm Hood

told me this one.

These representations are sneaky and probably cheating, but it is telling

you something important, namely, some representations for things will make

some problems dead easy, and others make it horribly di_cult. Thinking

about this early on can save you a lot of grief.

Exercise 2.5.5 Can you see why a parametric representation of the circle

of the form z = a+r cos _ +i(b+r sin _) could be a serious blunder in trying

to show that inversions take circles to circles?

Remark:

64 CHAPTER 2. EXAMPLES OF COMPLEX FUNCTIONS

The moral we draw from this little excursion is that being true and faithful

to a human being is, possibly, a _ne and splendid thing; being faithful and

true to a principle or ideology might be a _ne and splendid thing, or it might

be a sign of a sentimental nature gone wild. But being faithful and true to

a brand of beans or a choice of representation of an object is to confuse the

_nger pointing at the moon with the moon itself, and a sure sign of total

fatheadedness. The poor devil who believes deeply that the only true and

proper way to represent a circle in the plane is by writing down

(x 􀀀 a)2 + (y 􀀀b)2 = r2

is to be pitied as someone who has confused the language with the thing being

talked about, and is _t only for politics. The more ways you have of talking

and thinking about things, the easier it is to draw conclusions, and the harder

it is to be led astray. It is also a lot more fun.

The converse is also true: the inversion of a straight line is a circle through

the origin.

To see this, let ax + by + c = 0 be the equation of a straight line. Turn this

into polars to get

ar cos _ + br sin _ + c = 0

Now put r = 1=s to get the inversion:

(a=s) cos _ + (b=s) sin _ + c = 0

and rearrange to get

s2 + (as=c) cos _ + (bs=c) sin _ = 0

which is a circle passing through the origin with centre at _ 􀀀a=2c

􀀀b=2c _.

It is easy to see that the 'points at in_nity' on each end of the line get sent

to the origin.

This suggests that we could simplify the description by working not in the

plane but in the space we would get by adjoining a 'point at in_nity'.

We do this by putting a sphere of radius 1=2 sitting on the origin of R3 , and

identify the z = 0 plane with C . Now to map from the sphere to the plane,

take a line from the north pole of the sphere which is at the point2

4

0

0

1

35

2.5. THE FUNCTION F(Z) = 1

Z 65

P’

P

Q

Q’

Figure 2.21: The Riemann Sphere

and draw it so it cuts the sphere in P and the plane at P0. Now this sets up

a one-one correspondence between the points of the sphere other than the

north pole and the points of the plane. The unit circle in the plane is sent

to the equator of the sphere.

Now we put the 'point at in_nity' of the plane in- at the north pole of the

sphere.

An inversion of the plane now gives an inversion of the sphere, which sends

the South pole (the origin) to the North pole: all we do is to project down

so that the point Q goes directly to the point Q0 vertically below it, and

vice-versa. In other words, we reect in the plane of the equator.

Exercise 2.5.6 Verify that this rule ensures that a point in the plane is sent

to its inversion when we go from the point up to the sphere, then reect in

the plane of the equator, then go back to the plane.

Exercise 2.5.7 Suppose we have a disk which contains the origin on its

boundary. What would you expect the inversion of the disk to look like?

Suppose we have a disk which contains the origin in its interior. What would

you expect the inversion to look like?

66 CHAPTER 2. EXAMPLES OF COMPLEX FUNCTIONS

Sketches of the general situation should take you only a few minutes to work

out; it is probably easiest to visualise it on the Riemann Sphere.

Exercise 2.5.8 What would you expect to get, qualitatively, if you invert a

triangle shaped region of C ? Does it make a di_erence if the triangle contains

the origin?

Draw some pictures of some triangles and what you think their inversions

would look like.

Note that if you do an inversion and then invert the result, you get back to

where you started. In other words, the inversion is its own inverse map. Since

the same is true of conjugation, the map f(z) = 1=z also has this property.

Exercise 2.5.9 What happens if you invert a half-plane made by taking all

the points on one side of a line through the origin? What if the half-plane is

the set of points on one side of a line not through the origin?

I haven't said anything much about the conjugation because it is really very

trivial: just reect everything in the X-axis.

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